Ice In Water: Calculating Final Temperature In Calorimeter

Hey guys! Ever wondered what happens when you toss a piece of ice into a cup of water? It's not just about the ice melting; it's a whole dance of heat transfer and temperature changes! Let's dive into a classic physics problem involving a copper calorimeter, water, and ice to understand how we can calculate the final temperature of this icy mix. This problem is a fantastic example of thermodynamics in action, and we'll break it down step-by-step so it's super easy to follow. Understanding the principles behind this scenario can help you grasp more complex concepts in heat transfer and calorimetry. So, grab your thinking caps, and let's get started!

The Problem: An Icy Puzzle

Here’s the scenario: We have a copper calorimeter, which is basically a fancy insulated container used to measure heat changes. This calorimeter weighs 200 grams. Inside, there's 100 grams of water sitting at a cozy 16°C. Now, we introduce a 9.3 gram piece of ice, straight from the freezer at a chilly 0°C. The ice melts completely, and we want to figure out the final temperature of the water after everything has settled. Sounds like a cool puzzle, right? To tackle this, we'll use the principles of heat exchange and calorimetry, ensuring that we account for every gram of material and every degree of temperature change. The goal isn't just to find a number, but to understand the process by which heat is transferred and how it affects the final state of the system.

Breaking Down the Heat Exchange

The key to solving this lies in understanding that heat will flow from the warmer objects (the water and the calorimeter) to the colder object (the ice) until they all reach the same temperature. This is a fundamental principle of thermodynamics: heat always moves from a warmer body to a cooler one. To solve this, we need to consider a few things:

1. Heat gained by the ice to melt: Ice at 0°C needs to absorb heat to transform from a solid (ice) to a liquid (water). This heat is known as the latent heat of fusion. We'll need to calculate how much heat the ice absorbs just to melt.
2. Heat gained by the melted ice to warm up: Once the ice melts, it's water at 0°C. This water needs to warm up to the final temperature. We'll calculate the heat required for this temperature increase.
3. Heat lost by the water: The original 100 grams of water at 16°C will lose heat as it cools down to the final temperature.
4. Heat lost by the calorimeter: The copper calorimeter itself will also lose heat as it cools down along with the water. It's important to include this in our calculations because the calorimeter is part of the system exchanging heat.

By carefully accounting for each of these heat transfers, we can set up an equation that balances the heat gained and the heat lost. This balance will allow us to solve for the final temperature. The process involves using the specific heat capacities of water and copper, as well as the latent heat of fusion for ice. This ensures that we're accurately tracking the energy flow within the system.

Key Formulas and Concepts

Before we jump into the calculations, let's arm ourselves with the necessary formulas and concepts. These are the building blocks of our solution, and understanding them will make the process much clearer. Here are the key players:

• Heat (Q): This is the energy transferred due to temperature differences. We'll be calculating heat gained and heat lost. The standard unit for heat is Joules (J).
• Specific Heat Capacity (c): This is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (°C). Different materials have different specific heat capacities. For example, water has a higher specific heat capacity than copper, meaning it takes more energy to heat up water compared to copper. We'll need the specific heat capacity of water (${c_w}$) and copper (${c_c}$).
• Latent Heat of Fusion (L): This is the amount of heat required to change a substance from a solid to a liquid (or vice versa) at its melting point. For ice, this is the heat required to melt it into water at 0°C. We'll use the latent heat of fusion of ice (${L_f}$).
• Formula for Heat Transfer (Q = mcΔT): This is the main formula we'll use. It states that the heat (Q) gained or lost by a substance is equal to its mass (m) times its specific heat capacity (c) times the change in temperature (ΔT).
• Formula for Latent Heat (Q = mL): This formula calculates the heat required for a phase change (like melting). The heat (Q) is equal to the mass (m) times the latent heat of fusion (L).

With these formulas in our arsenal, we're well-equipped to tackle the heat exchange calculations. Remember, the core idea is to equate the total heat gained by the ice to the total heat lost by the water and the calorimeter. This principle of energy conservation is what allows us to solve for the final temperature.

Step-by-Step Calculation

Okay, let's put those formulas to work and crunch some numbers! This is where we get to see how the theory translates into a concrete solution. We'll go through each step carefully, making sure we understand what's happening at each stage of the heat exchange process.

1. Heat Gained by the Ice to Melt (${Q_1}$)

The first thing the ice does is melt. To calculate the heat required for this, we use the latent heat of fusion formula: ${Q_1 = m_{ice} imes L_f}$ Where:

• ${m_{ice} = 9.3 ext{ g} = 0.0093 ext{ kg}}$ (mass of the ice)
• ${L_f = 3.34 imes 10^5 ext{ J/kg}}$ (latent heat of fusion of ice)

Plugging in the values, we get: ${Q_1 = 0.0093 ext{ kg} imes 3.34 imes 10^5 ext{ J/kg} = 310.62 ext{ J}}$ So, the ice absorbs 310.62 Joules of heat just to melt into water at 0°C.

2. Heat Gained by the Melted Ice to Warm Up (${Q_2}$)

Now that the ice has melted, we have 9.3 grams of water at 0°C. This water needs to warm up to the final temperature (${T_f}$). We'll use the heat transfer formula: ${Q_2 = m_{ice} imes c_w imes (T_f - 0)}$ Where:

• ${m_{ice} = 0.0093 ext{ kg}}$ (mass of the melted ice)
• ${c_w = 4186 ext{ J/kg°C}}$ (specific heat capacity of water)
• ${T_f}$ is the final temperature (what we're trying to find)

So, ${Q_2 = 0.0093 ext{ kg} imes 4186 ext{ J/kg°C} imes T_f = 38.93 T_f ext{ J}}$

3. Heat Lost by the Water (${Q_3}$)

The original 100 grams of water at 16°C will lose heat as it cools down to the final temperature. Again, we use the heat transfer formula: ${Q_3 = m_w imes c_w imes (16 - T_f)}$ Where:

• ${m_w = 0.1 ext{ kg}}$ (mass of the water)
• ${c_w = 4186 ext{ J/kg°C}}$ (specific heat capacity of water)

So, ${Q_3 = 0.1 ext{ kg} imes 4186 ext{ J/kg°C} imes (16 - T_f) = 418.6 imes (16 - T_f) ext{ J}}$

4. Heat Lost by the Calorimeter (${Q_4}$)

The copper calorimeter also loses heat. We use the heat transfer formula one more time: ${Q_4 = m_c imes c_c imes (16 - T_f)}$ Where:

• ${m_c = 0.2 ext{ kg}}$ (mass of the calorimeter)
• ${c_c = 385 ext{ J/kg°C}}$ (specific heat capacity of copper)

So, ${Q_4 = 0.2 ext{ kg} imes 385 ext{ J/kg°C} imes (16 - T_f) = 77 imes (16 - T_f) ext{ J}}$

5. Balancing the Heat Exchange

Now comes the crucial part: balancing the heat. The total heat gained by the ice (to melt and warm up) must equal the total heat lost by the water and the calorimeter: ${Q_1 + Q_2 = Q_3 + Q_4}$ Substituting our calculated values: ${310.62 + 38.93 T_f = 418.6 imes (16 - T_f) + 77 imes (16 - T_f)}$

Solving for the Final Temperature

Alright, we've got our equation set up, now it's time to put on our algebra hats and solve for the final temperature (${T_f}$). This part involves a bit of manipulation, but don't worry, we'll break it down step-by-step so it's crystal clear.

Let's rewrite our equation: ${310.62 + 38.93 T_f = 418.6 imes (16 - T_f) + 77 imes (16 - T_f)}$

First, we'll expand the terms on the right side: ${310.62 + 38.93 T_f = (418.6 imes 16) - 418.6 T_f + (77 imes 16) - 77 T_f}$ ${310.62 + 38.93 T_f = 6697.6 - 418.6 T_f + 1232 - 77 T_f}$

Next, let's combine the constant terms and the ${T_f}$ terms on the right side: ${310.62 + 38.93 T_f = 7929.6 - 495.6 T_f}$

Now, we want to isolate ${T_f}$. Let's move all the ${T_f}$ terms to the left side and the constant terms to the right side: ${38.93 T_f + 495.6 T_f = 7929.6 - 310.62}$ ${534.53 T_f = 7618.98}$

Finally, we can solve for ${T_f}$ by dividing both sides by 534.53: {T_f = rac{7618.98}{534.53} } ${T_f hickapprox 14.25 ext{ °C}}$

So, after all that calculation, we find that the final temperature of the water and the calorimeter is approximately 14.25°C. How cool is that? This result shows how the heat from the warmer water and calorimeter was used to first melt the ice and then warm the resulting water, eventually reaching a thermal equilibrium.

Conclusion: The Final Chill

And there you have it! We've successfully calculated the final temperature of the water in the calorimeter after adding ice. This problem beautifully illustrates the principles of heat exchange, specific heat capacity, and latent heat of fusion. We saw how heat flows from warmer objects to cooler ones until equilibrium is reached, and how we can use mathematical equations to predict these temperature changes. The final temperature of approximately 14.25°C shows the impact of the ice cooling the system, but not by a huge amount, thanks to the relatively small mass of ice compared to the water and the calorimeter.

By breaking down the problem into smaller steps—calculating the heat required to melt the ice, the heat gained by the melted ice, and the heat lost by the water and calorimeter—we were able to apply the principle of energy conservation and solve for the unknown final temperature. These types of problems aren't just academic exercises; they help us understand how thermal systems work in the real world, from refrigerators to engines.

So, the next time you drop an ice cube into your drink, remember the fascinating physics at play! You're witnessing a mini-calorimetry experiment right in your glass. Keep exploring, keep questioning, and keep learning, guys! Physics is all around us, making the world a cooler place – literally!